Is this a valid way to show that the recursive sequence $x_n = x_{n-1} + \frac{1}{x_{n-1}^2}$ is unbounded? I'm working through some analysis textbooks on my own, so I don't want the full answer. I'm only looking for a hint on this problem.
Rosenlicht's Introduction to Analysis asks me to prove that $x_n = x_{n-1} + \frac{1}{x_{n-1}^2}$, where $x_1 = 1$, is unbounded. I'm not sure how to approach this, but here's what I tried. 
To save myself typing, I let $b := x_n$ and $a := x_{n-1}$. 
Since $b > a > 1$ as $n \to \infty$, we know that 
\begin{align}
b &> a \\
e^{\ln b} &> e^{\ln a} \\
\frac{e^{\ln b}}{e^{\ln a}} &> 1 \\
e^{\ln b - \ln a} &> 1
\end{align}
but I'm not sure where to go from here, or if this is even the right direction. I'm trying to show that $b - a > \ln b - \ln a$, because then I can say that the sequence is always growing faster than the logarithmic function, which I know is unbounded.
To show that $b - a > \ln b - \ln a$, I tried proof by contradiction. If $b - a \le \ln b - \ln a$, then $e^{b-a} \le e^ {\ln b - \ln a}$. Thus $\frac{e^b}{e^a} \le \frac{e^{\ln b}}{e^{\ln a}} = \frac{b}{a}$, and once again I'm not sure where to go. 
I know that to show something is bounded, I need to show that $\exists M > 0$ s.t. $x_n < M, \forall n$. I know how to do that with non-recursive sequences, e.g. $x_n = f(n), n \in \mathbb{N}$ because it's just algebra, but I'm not sure how to go about this with a recursive sequence (once that I wasn't successful at putting in non-recursive terms). 
 A: Hint #1: The sequence is monotone increasing.  If it were bounded, it would have a limit.  What property would that limit satisfy?
Hint #2: That limit would satisfy $$L=L+\frac{1}{L^2}$$
However no $L$ satisfies this equation.
A: Suppose $\{x_n\}$ is a bounded sequence, that is $0<x_n \leq M$ for some $M$. Then:
$x_n = (x_n-x_{n-1})+(x_{n-1}-x_{n-2})+\cdots +(x_2-x_1)+x_1 = \dfrac{1}{x_{n-1}^2}+\dfrac{1}{x_{n-2}^2}+ \cdots +\dfrac{1}{x_1^2}>\dfrac{1}{M^2}+\dfrac{1}{M^2}+\cdots +\dfrac{1}{M^2}=\dfrac{n-1}{M^2}$. If we select $n$ be such that $n > M^3+1$, then $x_n > M$, and it shows $\{x_n\}$ is not bounded.
A: New one on me. Notice that the same sort of thing happens if we are adding something really tiny, for example
$$ y_{n+1} = y_n + \frac{1}{e^{y_n}};  $$ it just takes much longer to pass any given point.
So I wanted to emphasize that, we say the sequence is unbounded, how long does it take to get to specified points? This is the original example,
 $$ x_{n+1} = x_n + \frac{1}{x_n^2}.  $$
Let us say that it takes one step to get us to $x_1 = 1.$
How long does it stay with $1 \leq x_n \leq 2?$ We are adding at least $1/4$ each time, so we reach or pass $2$ within four steps.
How long does it stay with $2 \leq x_n \leq 3?$ We are adding at least $1/9$ each time, so we reach or pass $3$ within another nine steps.
How long does it stay with $3 \leq x_n \leq 4?$ We are adding at least $1/16$ each time, so we reach or pass $4$ within another sixteen steps.
So, we are guaranteed to reach some positive integer $M$ within
$$ 1 + 4 + 9 + 16 + \cdots + M^2 = \frac{1}{6} M (M+1)(2M+1)   $$
steps. Hope I remembered the sum correctly.
The same sort of argument works for the $y_n$ I gave above, but instead of adding the integers squared each time we add $e^w$ for $w-1 \leq y_n \leq w,$ so the sum to get to some $M$ is quite huge, finite geometric sum which can be calculated.
Rosenlicht was a good guy. My friend Dmitry was his student. I think, after his retirement, Rosy went to teach for a few years in Kenya.  Dmitry says also the Ivory Coast. 
A: It's straightforward to show by induction that the sequence is strictly positive and strictly increasing. Therefore the sequence converges if and only if it is bounded above. Suppose there exists $M$ such that $x_n<M$ for all $n$. Then $\frac1{x_{n-1}}>\frac1M$ so $\frac1{x_{n-1}^2}>\frac1{M^2}$. But this implies that $x_n-x_{n-1}=\frac1{x_{n-1}^2}>\frac1{M^2}$ for all $n$, so $\{x_n\}$ cannot converge (take $\varepsilon <\frac1{M^2}$). This is a contradiction, so we conclude that $\{x_n\}$ is not bounded.
